dongzhao2725 2011-04-07 11:22
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PHP生成UL LI,UL LI

Can't figure out how-to generate this menu using a while-loop.

This is an example of my code:

<ul id="nav">
<li><a href="#">Hoofdmenu 1</a>
<ul class="sub">
        <li><a href="#">Submenu 1.1</a></li>
        <li><a href="#">Submenu 1.2</a></li>
        <li><a href="#">Submenu 1.3</a></li>
        <li><a href="#">Submenu 1.4</a></li>
    </ul>
</li>

<li><a href="#">Hoofdmenu 2</a>
    <ul class="sub">
        <li><a href="#">Submenu 2.1</a></li>
        <li><a href="#">Submenu 2.2</a></li>
        <li><a href="#">Submenu 2.3</a></li>
        <li><a href="#">Submenu 2.4</a></li>
    </ul>
</li>
</ul>

My dbtable looks like:

paginas:
    id
    title
    content
    type

When type == id from the parent it should be the submenu. In my example this works, now I've got to make it dynamic. Brains ain't working atm.

Thanks for your help!

Used code to get data from db:

<ul id="nav">
<?php
include_once("ond/inc/php/connect.php");
$query = "SELECT * FROM paginas WHERE type = '0'";
$result = mysql_query($query);
while($row = mysql_fetch_object($result)){

echo '<li><a href="?ond='.$row->titel.'">'.$row->titel.'</a>';}
echo '<ul class="sub">';

$query2 = "SELECT * FROM paginas WHERE type = '".$row->id."'";
$result2 = mysql_query($query2);    
while($row2 = mysql_fetch_object($result2))
{
    echo '<li><a href="?ond='.$row2->titel.'">'.$row2->titel.'</a></li>';
}
echo '</ul>'; 
echo '</li>';

?>
</ul>
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3条回答 默认 最新

  • doufei1852 2011-04-07 12:43
    关注

    Next lines did the solution:

    <ul id="nav">
    <?php
    include_once("ond/inc/php/connect.php");
    $query = "SELECT * FROM paginas WHERE type = '0'";
    $result = mysql_query($query);
    while($row = mysql_fetch_object($result)){
    
    echo '<li><a href="?ond='.$row->titel.'">'.$row->titel.'</a>';
    
    $query2 = "SELECT * FROM paginas WHERE type = '".$row->id."'";
    $result2 = mysql_query($query2);    
    echo '<ul class="sub">';
    while($row2 = mysql_fetch_object($result2))
    {   
        echo '<li><a href="?ond='.$row2->titel.'">'.$row2->titel.'</a></li>';
    
    
    }
        echo '</ul>';
    echo '</li>';}
    
    
    
    ?>
    </ul>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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